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I have read that an operation can have at most one identity and fully understand the proof, however what if I define an operation $*$ on $\mathbb Q$ as follows?

$x*y=|x \times y|$, $\forall x,y \in \mathbb Q$

Surely both $1$ and $-1$ are identities? Or am I missing something obvious?

JMP
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RedG
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  • What does $xy$ mean for an abstract set? If, say, you meant to assume that $S\in \mathbb Q^$ or such, then using your operation $-1-1=1$, so $-1$ is not an identity. – lulu Oct 01 '16 at 11:24
  • You can't have more than one identity because $e_1\star e_2$ would have to be both $e_1$ and $e_2$. – lulu Oct 01 '16 at 11:25
  • If $x\lt0$ then the equation $x*y=x$ has no solution, so your operation has no neutral element at all. – bof Oct 01 '16 at 11:31

1 Answers1

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Note that neither $1$ nor $-1$ is an identity. We have that:

$$(-1)*(-1) = \;\mid 1 \mid\; = 1 \quad \quad (-1)*(1) = \;\mid -1 \mid\; = 1$$

Stefan4024
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